Hi,
Is possible to use functions in conditional operator as ashown below.
.....
.....
int fun1()
{
// body
}
int fun2()
{
//body
}
....
main()
{...
int x;
x = (a>b)?fun1():fu n2();
}
30  2172 
"bnp" <bh**********@y ahoo.com> a écrit dans le message de
news:7d******** *************** **@posting.goog le.com... Hi,
Hi,
Is possible to use functions in conditional operator as ashown below.
....
....
int fun1()
{
// body
}
int fun2()
{
//body
}
...
main()
{...
int x;
x = (a>b)?fun1():fu n2();
Yes,
it's semantically equivalent to:
if (a>b)
x = fun1();
else
x = fun2();
IMHO, I find that the conditional operator fits very well in this case.
Regis
In message <7d************ *************@p osting.google.c om> bh**********@ya hoo.com (bnp) wrote: Hi,
Is possible to use functions in conditional operator as ashown below.
x = (a>b)?fun1():fu n2();
Yes. And even more entertainingly, you can do this:
x = (a > b ? fun1 : fun2) (c, d, e);
--
Kevin Bracey, Principal Software Engineer
Tematic Ltd Tel: +44 (0) 1223 503464
182-190 Newmarket Road Fax: +44 (0) 1728 727430
Cambridge, CB5 8HE, United Kingdom WWW: http://www.tematic.com/
Kevin Bracey wrote: Yes. And even more entertainingly, you can do this:
x = (a > b ? fun1 : fun2) (c, d, e);
Yes, of course! Never thought of stretching the meaning
of a function 'pointer' this far. Things like 5["Hello world"]
(as you know I'm sure) are entertaining too.
Kees
Kevin Bracey <ke**********@t ematic.com> wrote:
: In message <7d************ *************@p osting.google.c om>
: bh**********@ya hoo.com (bnp) wrote:
:> Hi,
:>
:> Is possible to use functions in conditional operator as ashown below.
:> x = (a>b)?fun1():fu n2();
: Yes. And even more entertainingly, you can do this:
: x = (a > b ? fun1 : fun2) (c, d, e);
Are you saying that if functions 'fun1' and 'fun2' have the same
prototypes and take (appropriately typed) arguments, 'c', 'd', and 'e',
that this would be equivalent to:
if ( a > b )
x = fun1( c, d, e );
else
x = fun2( c, d, e );
?
Paul
--
Paul D. Boyle bo***@laue.chem .ncsu.edu
North Carolina State University http://www.xray.ncsu.edu
Kevin Bracey wrote: Yes. And even more entertainingly, you can do this:
x = (a > b ? fun1 : fun2) (c, d, e);
BTW what other seemingly strange standard C construction
do you know? I mentioned 5["fsdssdfssf "] which looks
very strange to most people even very experienced C
programmers. It would be fun to discuss this kind of
constructs.
Case
bnp wrote: Hi,
Is possible to use functions in conditional operator as ashown below.
....
....
int fun1()
{
// body
}
int fun2()
{
//body
}
...
main()
{...
int x;
x = (a>b)?fun1():fu n2();
}
Yes. You can even use `x = (a > b ? fun1 : fun2)();'
if you have sado-masochistic tendencies.
-- Er*********@sun .com
Case wrote: Kevin Bracey wrote:
Yes. And even more entertainingly, you can do this:
x = (a > b ? fun1 : fun2) (c, d, e);
BTW what other seemingly strange standard C constructions
do you know? I mentioned 5["fsdssdfssf "] which looks
very strange to most people
About the whole human race of course. But I meant C
programmers. :-)
Kees
bnp wrote:
(snip) Is possible to use functions in conditional operator as ashown below.
(snip)
int x;
x = (a>b)?fun1():fu n2();
Not only that, you can even do
x=(a>b?fun1:fun 2)();
assuming that fun1 and fun2 are declared as functions.
(It is more interesting if they have arguments.)
-- glen
Paul D. Boyle wrote:
Kevin Bracey <ke**********@t ematic.com> wrote:
: In message <7d************ *************@p osting.google.c om>
: bh**********@ya hoo.com (bnp) wrote:
:> Hi,
:>
:> Is possible to use functions in conditional operator as ashown below.
:> x = (a>b)?fun1():fu n2();
: Yes. And even more entertainingly, you can do this:
: x = (a > b ? fun1 : fun2) (c, d, e);
Are you saying that if functions 'fun1' and 'fun2' have the same
prototypes and take (appropriately typed) arguments, 'c', 'd', and
'e',
It could happen. Take a look at this line of code
string = (digit == 0 ? sput_ip1 : sput_i)(integer / 10, string);
from http://groups.google.com/groups?selm...mindspring.com
sput_ip1 and sput_i, are functions.
--
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